Answer
$$y = x - 1$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^3}\ln x,{\text{ }}\left( {1,0} \right) \cr
& {\text{Differentiate }}f\left( x \right) \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^3}\ln x} \right] \cr
& f'\left( x \right) = {x^3}\left( {\frac{1}{x}} \right) + \ln x\left( {3{x^2}} \right) \cr
& f'\left( x \right) = {x^2} + 3{x^2}\ln x \cr
& {\text{Calculate the slope at the given point }}\left( {1,0} \right) \cr
& m = f'\left( 1 \right) = {\left( 1 \right)^2} + 3{\left( 1 \right)^2}\ln \left( 1 \right) \cr
& m = 1 \cr
& {\text{Find the equation of the tangent line at }}\left( {1,0} \right) \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - 0 = \left( {x - 1} \right) \cr
& y = x - 1 \cr
& \cr
& {\text{Graph}} \cr} $$
